# When is the converse of Casorati–Weierstrass false?

Recall that Casorati-Weierstrass theorem says that given $f$ is holomorphic in the punctured disc $D_r(z_0) - \{z_0\}$ and has an essential singularity at $z_0$. Then, the image of $D_r(z_0) - \{z_0\}$ under $f$ is dense in the complex plane.

When does the converse fail, if ever? It seems like IF the image of $f$ around a neighborhood of $z_0$ is dense in the complex plane, then neither the limit of $f$ will exist as $z \rightarrow z_0$, nor will it be unbounded because then it would be impossible for me to exhibit convergent sequences as $z\rightarrow z_0$. So what am I missing/misunderstanding?

• I'd say the converse is the trivial direction. If the function has a removable singularity at a point, it's obviously bounded near it, so the image can't be dense. Similarly, if the function has a pole then its modulus must be bounded away from zero and the image will now miss the nbhd of 0. I suspect this is why the theorem is not stated as an equivalence. Commented Jan 1, 2014 at 11:30
• @Marek Thank you for the remark. Commented Jan 1, 2014 at 11:43
• Careful. If you're looking at only one $r > 0$, it can easily be that $f(D_r(z_0)\setminus\{z_0\})$ is dense in the plane but $z_0$ is a removable singularity or a pole. You need that $f(D_r(z_0)\setminus\{z_0\})$ is dense for all $r > 0$ (small enough that $f$ is defined on the punctured disk). Commented Jan 1, 2014 at 12:50
• @DanielFischer, thanks. Were you thinking of, say for $z_0 = 0$ and $r_0$ real, $f(re^{i\theta}) = \exp(\frac{1}{r-r_0}e^{i\theta})$? Commented Jan 1, 2014 at 16:18
• No. That doesn't look holomorphic at first glance, by the way. I was thinking of something like $f = \lambda \circ T$, where $\lambda$ is the modular function on the upper half plane, and $T$ is a Möbius transformation mapping the disk onto the upper half plane. Then $f(D_r(z_0)\setminus\{z_0\}) = \mathbb{C}\setminus \{0,1\}$, and you only get a non-dense image when you restrict to a radius $\rho < r$. Commented Jan 1, 2014 at 16:26

EDIT: In fact there is a technicality here; we have to assume that $$f$$ is defined on the closed disc, or the behavior of $$f$$ on the boundary comes into play.